Therefore, the angular acceleration is 3.925 rad/s2.

[gcombina]

Homework Statement

A grindstone, initially at rest, is given a constant angular acceleration so that it makes 20.0 rev in the first 8.00 s. What is its angular acceleration?
(a) 0.313 rad/s2 (c) 2.50 rad/s2 (e) 3.93 rad/s2

(b) 0.625 rad/s2 (d) 1.97 rad/s2

Homework Equations

The Attempt at a Solution

#1. I am changing the rev part to radians

20rev = 125.66

#2. I am using the formula

w^2 = w^2 + 2 alpha tetha (where are the symbols?)

 

[Dr.D]

What do you mean when you write

"w^2 = w^2 + 2 alpha tetha "

If we subtract w^2 from both sides, doesn't this leave you with 2 alpha tetha = ? Where did you get that last formula?

By the way, I think you most likely mean theta, rather than tetha.

 

[gcombina]

The equation you used relates the acceleration, displacement, initial and final velocities, and leaves out the time. If the question asked to find the angular acceleration given : that the wheel starts from rest (ωo = 0), and reaches a velocity of ωf rad/sec when it is stopped after making Θ number of revolutions, you would use ω^2 - ωo^2 = 2∝ Θ
The equation you want is the one that includes the time t.
***** Θ = Vo t + 1/2 (∝) (t^2) ******

Θ = 20 rev x (2 pi rad/ rev) = 125.6 tad

125.6 = 0 + 0.5 (or 1/2) ∝ (8)^2 and
∝ = 125.6/32
= 3.925 rad/sec^2